A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{4}\]
C) \[-\frac{\pi }{2}\]
D) None of these
Correct Answer: B
Solution :
Since, \[\frac{d}{dx}{{\tan }^{-1}}\left( \frac{1}{x} \right)=\frac{d}{dx}{{\cot }^{-1}}x=-\frac{1}{1+{{x}^{2}}}\] \[\therefore \]\[\int_{-1}^{1}{\frac{d}{dx}\left( {{\tan }^{-1}}\left( \frac{1}{x} \right) \right)}dx\] \[=-\int_{-1}^{1}{\frac{1}{1+{{x}^{2}}}}dx\] \[=-2\int_{0}^{1}{\frac{1}{1+{{x}^{2}}}}dx=-2[{{\tan }^{-1}}x]_{0}^{1}\] \[=-\frac{\pi }{2}\]
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